BZOJ 2820. YY的GCD

 

$$\sum_{i=1}^{n}\sum_{i=1}^{m}\left[\left(i,j\right)\in P\right]$$
$$=\sum_{p\in P}\sum_{i=1}^{n}\sum_{i=1}^{m}\left[\left(i,j\right)=p\right]$$$$=\sum_{p\in P}\sum_{i=1}^{\lfloor \frac{n}{p} \rfloor}\sum_{i=1}^{\lfloor \frac{m}{p} \rfloor}\left[\left(i,j\right)=1\right]$$
$$=\sum_{p\in P}\sum_{d=1}^{\min\{\lfloor \frac{n}{p} \rfloor,\lfloor \frac{m}{p} \rfloor\}}\mu(d)\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor$$
$$=\sum_{T=1}^{\min\{n, m\}}\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum_{p|T \wedge p\in P}\mu(\frac{T}{p})$$

$f(n) = \sum_{p|n \wedge p \in P}\mu(\frac{n}{p})$ 不是积性函数，但是可以线性筛预处理出来。
对于质数 $p$，$f(p)=1$
当 $p < M(i)$ 时，$f(i \times p)=-f(i)+\mu(i)$
当 $p=M(i)$ 时，$i \times p$ 里含有 $p^2$，所以 $\forall p' \neq p$，$\mu(\frac{i\times p}{p'})=0$，所以 $f(i\times p)=\mu(i)$

#include <bits/stdc++.h>
#define ll long long

const int N = 1e7 + 7;
int prime[N], prin, mu[N];
bool vis[N];
ll f[N];

void init() {
    mu[1] = 1;
    for (int i = 2; i < N; i++) {
        if (!vis[i]) {
            prime[++prin] = i;
            mu[i] = -1;
            f[i] = 1;
        }
        for (int j = 1; j <= prin && i * prime[j] < N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                f[i * prime[j]] = mu[i];
                break;
            }
            mu[i * prime[j]] = -mu[i];
            f[i * prime[j]] = -f[i] + mu[i];
        }
    }
    for (int i = 1; i < N; i++)
        f[i] += f[i - 1];
}

ll solve(int n, int m) {
    ll ans = 0;
    for (int i = 1, j; i <= std::min(n, m); i = j + 1) {
        j = std::min(n / (n / i), m / (m / i));
        ans += (f[j] - f[i - 1]) * (n / i) * (m / i);
    }
    return ans;
}

int main() {
    init();
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        printf("%lld\n", solve(n, m));
    }
    return 0;
}